Section 10.6: The Inverse Trigonometric Functions - Ostts.org

836 Foundations of Trigonometry2. (a) Since the domain of F (x) = arccos(x) is −1 ≤ x ≤ 1, we can find the domain off(x) = π 2 − arccos ( )x5 by setting the argument of the arccosine, in this casex5 , between−1 and 1. Solving −1 ≤ x 5≤ 1 gives −5 ≤ x ≤ 5, so the domain is [−5, 5]. To determinethe range of f, we take a cue from Section 1.7. Three ‘key’ points on the graph ofF (x) = arccos(x) are (−1, π), ( 0, π )2 and (1, 0) . Following the procedure outlined inTheorem 1.7, we track these points to ( −5, − π ) ( )2 , (0, 0) and 5,π2 . Plotting these valuestells us that the range 5 of f is [ − π 2 , π ]2 . Our graph confirms our results.(b) To find the domain and range of f(x) = 3 arctan (4x), we note that since the domainof F (x) = arctan(x) is all real numbers, the only restrictions, if any, on the domain off(x) = 3 arctan (4x) come from the argument of the arctangent, in this case, 4x. Since4x is defined for all real numbers, we have established that the domain of f is all realnumbers. To determine the range of f, we can, once again, appeal to Theorem 1.7.Choosing our ‘key’ point to be (0, 0) and tracking the horizontal asymptotes y = − π 2and y = π 2, we find that the graph of y = f(x) = 3 arctan (4x) differs from the graph ofy = F (x) = arctan(x) by a horizontal compression by a factor of 4 and a vertical stretchby a factor of 3. It is the latter which affects the range, producing a range of ( − 3π 2 , 3π )2 .We confirm our findings on the calculator below.y = f(x) = π 2 − arccos ( x5)y = f(x) = 3 arctan (4x)(c) To find the domain of g(x) = arccot ( )x2 + π, we proceed as above. Since the domain ofG(x) = arccot(x) is (−∞, ∞), and x 2is defined for all x, we get that the domain of g is(−∞, ∞) as well. As for the range, we note that the range of G(x) = arccot(x), like thatof F (x) = arctan(x), is limited by a pair of horizontal asymptotes, in this case y = 0and y = π. Following Theorem 1.7, we graph y = g(x) = arccot ( )x2 + π starting withy = G(x) = arccot(x) and first performing a horizontal expansion by a factor of 2 andfollowing that with a vertical shift upwards by π. This latter transformation is the onewhich affects the range, making it now (π, 2π). To check this graphically, we encountera bit of a problem, since on many calculators, there is no shortcut button correspondingto the arccotangent function. Taking a cue from number 1c, we attempt to rewriteg(x) = arccot ( )x2 +π in terms of the arctangent function. Using Theorem 10.27, we havethat arccot ( ) (x2 = arctan 2)x whenx2> 0, or, in this case, when x > 0. Hence, for x > 0,we have g(x) = arctan ( )2x + π. Whenx2< 0, we can use the same argument in number1c that gave us arccot(−2) = π + arctan ( − 1 (2)to give us arccotx) (2 = π + arctan 2x).5 It also confirms our domain!